JEE MAIN - Physics (2025 - 7th April Morning Shift - No. 19)
A cubic block of mass $m$ is sliding down on an inclined plane at $60^{\circ}$ with an acceleration of $\frac{g}{2}$, the value of coefficient of kinetic friction is
$\frac{\sqrt{3}}{2}$
$\frac{\sqrt{2}}{3}$
$1-\frac{\sqrt{3}}{2}$
$\sqrt{3}-1$
Explicació
_7th_April_Morning_Shift_en_19_1.png)
$$\begin{aligned} & m g \sin 60^{\circ}-\mu m g \cos 60^{\circ}=m a \\ & g \sin 60-\mu g \cos 60=\frac{g}{2} \\ & \frac{\sqrt{3}}{2}-\frac{\mu}{2}=\frac{1}{2} \\ & \mu=\sqrt{3}-1 \end{aligned}$$